Cauchy Product Radius of Convergence How do we see that the radius of convergence (RCV) of the Cauchy product is at least the minimum of the two respective RCV's?
For instance suppose $f(x)=\sum_{i=0}^\infty a_i(x-x_0)^i$ with RCV $R_1$.
$g(x)=\sum_{i=0}^\infty b_i(x-x_0)^i$ with RCV $R_2$.
What I know is since the two functions converge absolutely we may use Merten's Theorem to conclude that $fg(x)=\sum_{i=0}^\infty c_i(x-x_0)^i$, where $c_i=\sum_{j=0}^i a_jb_{i-j}$.
However I don't see how to see the RCV of the Cauchy product.
Thanks for any help!
 A: Hint. Let $0< r< \min(R_a,R_b)$. One may write
$$
\begin{align}
|c_n \:r^n|&=\left|\sum_{j=0}^n a_jr^jb_{n-j}r^{n-j}\right|
\\\\&\le \sum_{j=0}^n |a_jr^j|\:|b_{n-j}r^{n-j}|
\\\\&\le \left(\sum_{j=0}^n |a_jr^j|\right)\cdot \left(\sum_{j=0}^n |b_{n-j}r^{n-j}|\right)
\\\\&\le \left(\sum_{j=0}^\infty |a_jr^j|\right)\cdot \left(\sum_{j=0}^\infty |b_{n-j}r^{n-j}|\right)
\\\\&<\infty
\end{align}
$$ meaning the sequence $\left\{|c_n \:r^n|\right\}_{\mathbb{N}}$ is bounded, for any $|z|<r$, 
$$
|c_n \:z^n|=\left|c_nr^n \frac{z^n}{r^n} \right|\le M \left|\frac{z}{r}\right|^n
$$
thus $R_c\ge \min(R_a,R_b)$.
A: This is a straightforward outcome of Mertens Theorem, which states that if we have two infinite convergent series and at least one of them converges absolutely, then their Cauchy product also converges .
Since the convergence of power series is absolute within the convergence interval, we can apply the above theorem to any point in the smallest interval which is certainly contained in both series convergence interval.
